The solubility product (Ksp) of barium sulfate, BaSO4, at 25\u00b0C is given as 1.1 x 10^-10. We can use this value to calculate the solubility of BaSO4 in pure water in both moles per liter and grams per liter.<\/p>\n\n\n\n
Step 1: Write the dissociation equation for BaSO4<\/h3>\n\n\n\n
When BaSO4 dissolves in water, it dissociates according to the equation: BaSO4(s)\u21ccBa2+(aq)+SO42\u2212(aq)\\text{BaSO}_4 (s) \\rightleftharpoons \\text{Ba}^{2+} (aq) + \\text{SO}_4^{2-} (aq)BaSO4\u200b(s)\u21ccBa2+(aq)+SO42\u2212\u200b(aq)<\/p>\n\n\n\n
The stoichiometry of the dissociation tells us that for every mole of BaSO4 that dissolves, one mole of Ba\u00b2\u207a and one mole of SO\u2084\u00b2\u207b are produced.<\/p>\n\n\n\n
Step 2: Set up the expression for Ksp<\/h3>\n\n\n\n
The solubility product expression for BaSO4 is: Ksp=[Ba2+][SO42\u2212]K_{sp} = [\\text{Ba}^{2+}][\\text{SO}_4^{2-}]Ksp\u200b=[Ba2+][SO42\u2212\u200b]<\/p>\n\n\n\n
Let the solubility of BaSO4 be s<\/strong> moles per liter. At equilibrium: [Ba2+]=sand[SO42\u2212]=s[\\text{Ba}^{2+}] = s \\quad \\text{and} \\quad [\\text{SO}_4^{2-}] = s[Ba2+]=sand[SO42\u2212\u200b]=s<\/p>\n\n\n\n Therefore, the Ksp expression becomes: Ksp=s2K_{sp} = s^2Ksp\u200b=s2<\/p>\n\n\n\n Given that Ksp=1.1\u00d710\u221210K_{sp} = 1.1 \\times 10^{-10}Ksp\u200b=1.1\u00d710\u221210, we can substitute this value into the equation: 1.1\u00d710\u221210=s21.1 \\times 10^{-10} = s^21.1\u00d710\u221210=s2<\/p>\n\n\n\n Solving for s<\/strong>: s=1.1\u00d710\u221210=1.048\u00d710\u22125\u2009mol\/Ls = \\sqrt{1.1 \\times 10^{-10}} = 1.048 \\times 10^{-5} \\, \\text{mol\/L}s=1.1\u00d710\u221210\u200b=1.048\u00d710\u22125mol\/L<\/p>\n\n\n\n So, the solubility of BaSO4 in pure water is approximately 1.05\u00d710\u22125\u2009mol\/L1.05 \\times 10^{-5} \\, \\text{mol\/L}1.05\u00d710\u22125mol\/L.<\/p>\n\n\n\n Now, to convert the solubility from moles per liter to grams per liter, we use the molar mass of BaSO4. The molar mass is calculated as: Molar mass of BaSO4=137.33\u2009(Ba)+32.07\u2009(S)+4\u00d716.00\u2009(O)=233.39\u2009g\/mol\\text{Molar mass of BaSO}_4 = 137.33 \\, (\\text{Ba}) + 32.07 \\, (\\text{S}) + 4 \\times 16.00 \\, (\\text{O}) = 233.39 \\, \\text{g\/mol}Molar mass of BaSO4\u200b=137.33(Ba)+32.07(S)+4\u00d716.00(O)=233.39g\/mol<\/p>\n\n\n\n Now, we can calculate the solubility in grams per liter: Solubility (g\/L)=s\u00d7Molar mass=(1.05\u00d710\u22125\u2009mol\/L)\u00d7233.39\u2009g\/mol=2.45\u00d710\u22123\u2009g\/L\\text{Solubility (g\/L)} = s \\times \\text{Molar mass} = (1.05 \\times 10^{-5} \\, \\text{mol\/L}) \\times 233.39 \\, \\text{g\/mol} = 2.45 \\times 10^{-3} \\, \\text{g\/L}Solubility (g\/L)=s\u00d7Molar mass=(1.05\u00d710\u22125mol\/L)\u00d7233.39g\/mol=2.45\u00d710\u22123g\/L<\/p>\n\n\n\n This calculation shows how to determine both the molar and mass solubility of BaSO4 using its solubility product (Ksp).<\/p>\n\n\n\n The Ksp for BaSO4 is 1.1×10^-10 at 25 \u00c2\u00b0C. Calculate the solubility of barium sulfate in pure water in moles per liter and grams per liter. The Correct Answer and Explanation is: The solubility product (Ksp) of barium sulfate, BaSO4, at 25\u00b0C is given as 1.1 x 10^-10. We can use this value to calculate […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-266389","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/266389","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=266389"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/266389\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=266389"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=266389"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=266389"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Step 3: Solve for s<\/h3>\n\n\n\n
Step 4: Convert the solubility to grams per liter<\/h3>\n\n\n\n
Final Answer:<\/h3>\n\n\n\n
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